A particle moves along the x-axis and has its displacement x varying with time t according to the equation x=c_0(t^2-2)+c(t-2)^2$x=c_{0}(t^{2}-2)+c(t-2)^{2}$ where c_0$c_{0}$ and c are constants of appropriate dimensions. Then, which of the following statements is correct?

### Related Formula The relationship between position x$x$, velocity v$v$, and acceleration a$a$ is given by differentiation with respect to time t$t$: v = fracdxdt$$v = \frac{dx}{dt}$$ a = fracdvdt$$a = \frac{dv}{dt}$$ ### Core Logic Given position: x = t^3 - 6t^2 + 20t + 15$$x = t^3 - 6t^2 + 20t + 15$$ Differentiating once to find velocity v$v$: v = fracdxdt = 3t^2 - 12t + 20$$v = \frac{dx}{dt} = 3t^2 - 12t + 20$$ Differentiating again to find acceleration a$a$: a = fracdvdt = 6t - 12$$a = \frac{dv}{dt} = 6t - 12$$ ### Step 1: Determine Time when Acceleration is Zero Set the acceleration to zero: a = 0 implies 6t - 12 = 0 implies t = 2text seconds$$a = 0 \implies 6t - 12 = 0 \implies t = 2\text{ seconds}$$ So, the acceleration becomes zero at t = 2text s$t = 2\text{ s}$. ### Step 2: Calculate Velocity at this Time Substitute t = 2text s$t = 2\text{ s}$ into the velocity equation: v = 3(2)^2 - 12(2) + 20$$v = 3(2)^2 - 12(2) + 20$$ v = 3(4) - 24 + 20$$v = 3(4) - 24 + 20$$ v = 12 - 24 + 20 = 8text m/s$$v = 12 - 24 + 20 = 8\text{ m/s}$$ Thus, the velocity is 8text m/s$8\text{ m/s}$. ### Pattern Recognition Sees: displacement function of degree 3 → acceleration is linear in time. The zero of a linear function of form At - B = 0$At - B = 0$ is easily calculated, and substituting back resolves to a basic quadratic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line

### Related Formula For a body starting from rest (u = 0$u = 0$) with constant acceleration a$a$, displacement S$S$ in time t$t$ is: S = frac12 a t^2$$S = \frac{1}{2} a t^2$$ ### Core Logic Let the constant acceleration be a$a$. * Displacement S_1$S_1$ covered in the first (p - 1)$(p - 1)$ seconds: S_1 = frac12 a (p - 1)^2$$S_1 = \frac{1}{2} a (p - 1)^2$$ * Displacement S_2$S_2$ covered in the first p$p$ seconds: S_2 = frac12 a p^2$$S_2 = \frac{1}{2} a p^2$$ ### Step 1: Express Net Displacement Let the time taken to achieve a displacement of S_1 + S_2$S_1 + S_2$ be t$t$. S_1 + S_2 = frac12 a t^2$$S_1 + S_2 = \frac{1}{2} a t^2$$ Substituting the expressions of S_1$S_1$ and S_2$S_2$: frac12 a (p - 1)^2 + frac12 a p^2 = frac12 a t^2$$\frac{1}{2} a (p - 1)^2 + \frac{1}{2} a p^2 = \frac{1}{2} a t^2$$ ### Step 2: Solve for Time Dividing the entire equation by frac12a$\frac{1}{2}a$: (p - 1)^2 + p^2 = t^2$$(p - 1)^2 + p^2 = t^2$$ p^2 - 2p + 1 + p^2 = t^2$$p^2 - 2p + 1 + p^2 = t^2$$ t^2 = 2p^2 - 2p + 1$$t^2 = 2p^2 - 2p + 1$$ t = sqrt2p^2 - 2p + 1 mathrm~s$$t = \sqrt{2p^2 - 2p + 1} \mathrm{~s}$$ ### Pattern Recognition When dealing with equations of motion from rest, notice that displacement scales quadratically with time (S propto t^2$S \propto t^2$). This means if displacement sums up (S_t = S_1 + S_2$S_t = S_1 + S_2$), the corresponding times will add in quadrature: t = sqrtt_1^2 + t_2^2$t = \sqrt{t_1^2 + t_2^2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line

### Related Formula v = u + at$v = u + at$ s = ut + frac12at^2$$s = ut + \frac{1}{2}at^2$$ S_n^textth = u + fraca2(2n - 1)$$S_{n^{\text{th}}} = u + \frac{a}{2}(2n - 1)$$ ### Core Logic Let the velocity at time t$t$ be u$u$. The time interval Delta t = (t+1) - t = 1 mathrm~s$\Delta t = (t+1) - t = 1 \mathrm{~s}$. The increase in velocity over 1 second is exactly the acceleration a$a$. The displacement in that 1-second interval acts as the (t+1)^textth$(t+1)^{\text{th}}$ second displacement equation. ### Step 1: Determine Acceleration Increase in velocity Delta v = 50 mathrm~m/s$\Delta v = 50 \mathrm{~m/s}$ in 1 mathrm~s$1 \mathrm{~s}$. v = u + at$v = u + at$ u + 50 = u + a(1) Rightarrow a = 50 mathrm~m/s^2$$u + 50 = u + a(1) \Rightarrow a = 50 \mathrm{~m/s^2}$$ ### Step 2: Determine Velocity 'u' at time t Displacement in the 1-second interval from t$t$ to t+1$t+1$ is 125 mathrm~m$125 \mathrm{~m}$. Using s = ut' + frac12at'^2$s = ut' + \frac{1}{2}at'^2$ where t' = 1 mathrm~s$t' = 1 \mathrm{~s}$: 125 = u(1) + frac12a(1)^2$$125 = u(1) + \frac{1}{2}a(1)^2$$ 125 = u + frac502$$125 = u + \frac{50}{2}$$ 125 = u + 25 Rightarrow u = 100 mathrm~m/s$$125 = u + 25 \Rightarrow u = 100 \mathrm{~m/s}$$ ### Step 3: Distance in the next second We need the distance travelled in the (t+2)^textth$(t+2)^{\text{th}}$ second, which corresponds to the 1-second interval starting with an initial velocity equal to the velocity at t+1$t+1$. Alternatively, we can use the n^textth$n^{\text{th}}$ second formula directly by re-indexing. The velocity at start of this interval is u_new = u + a = 100 + 50 = 150 mathrm~m/s$u_{new} = u + a = 100 + 50 = 150 \mathrm{~m/s}$. Distance S_1textst$S_{1\text{st}}$ using new parameters: s = 150(1) + frac12(50)(1)^2 = 150 + 25 = 175 mathrm~m$$s = 150(1) + \frac{1}{2}(50)(1)^2 = 150 + 25 = 175 \mathrm{~m}$$ Or using PDF logic from base u$u$ with n=2$n=2$ (since n=1$n=1$ was t+1^textth$t+1^{\text{th}}$): S_2^textnd = u + fraca2[2n - 1] = 100 + 25[4 - 1] = 100 + 75 = 175 mathrm~m$$S_{2^{\text{nd}}} = u + \frac{a}{2}[2n - 1] = 100 + 25[4 - 1] = 100 + 75 = 175 \mathrm{~m}$$ ### Pattern Recognition For 1-second interval mechanics, Delta v$\Delta v$ directly yields a$a$. The displacement equation simplifies gracefully to s = u_start + a/2$s = u_{start} + a/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line